what is the wavelength of the photon released when an electron in a hydrogen atom transitions from = 6 to = 1?

Answer:

60 N.C

Explanation:

The box will move in the direction of the push and pull with a force of 60 N. C.

If you and your friend are pushing a box at the same constant speed. You pushed 20 feet in 10 minutes, your friend was pushing for 5 minutes. , then the distance must she travel so the two of you can travel at the same speed would be 30 feet.

The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object.

As given in the problem If you and your friend are pushing a box at the same constant speed. You pushed 20 feet in 10 minutes, your friend was pushing for 5 minutes.

The distance traveled by box = 20 + 10

                                                 = 30 feets

Thus, the distance must she travel so the two of you can travel at the same speed would be 30 feet.

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When a wire carrying a 26-a current travels perpendicular to the field of a powerful magnet, it experiences a 2.16-n force on the 4.00 cm of wire in the field. The average field strength is 1.6T

Less than 30 microteslas (0.3 gauss), which corresponds to most of South America and South Africa, to more than 60 microteslas (0.6 gauss), which corresponds to northern Canada, south of Australia, and parts of Siberia, are the strengths of the fields at the Earth's surface.

The magnitude of the wire whose length is L carrying a current I and making an angle θ with a magnetic field of density B is calculated using the following formula.

B = F/(I * L * Sin(θ)

Given that F = 2.16N

I = 30 A

L = 4cm = 4 * 10^-2m

θ = π/2

By substitution

B = 2.16/(30 * 4 * 10^-2 * Sin(π/2))

B = 1.6T

Hence, the average field strength is 1.6T

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The sound intensity level would increase by 20 decibels if 100 people shout instead of one. Hence option B is correct.

The sound intensity level (SIL) increases by 10*log(N), where N is the number of people shouting. With the help of this equation, we can determine the rise in SIL as follows: N₁ = 1 (one person shouting) and N₂ = 100 (100 people shouting) simultaneously,

ΔSIL = 10log(N₂/N₁)

ΔSIL = 10log(100/1)

ΔSIL = 10*2

ΔSIL = 20 dB

Therefore, the answer is B that says "The sound is 20 decibels higher".

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Answer:

-2 kg x m/s

Explanation:

-17 + 54 = 35 + x

x = -2

The angular acceleration (α) of the pulley is 0.383 rad/s², and the linear acceleration of the bucket is 0.0867 m/s². At t = 3.00 s, the angular velocity (ω) of the pulley is 1.15 rad/s, and the linear velocity (v) of the bucket is 0.260 m/s.

To find the angular acceleration (α) of the pulley, we can use the torque equation: τ = Iα, where τ is the torque and I is the moment of inertia. The torque is given by the frictional torque at the axle, so we have τ = 1.10 N·m. Rearranging the equation, we get α = τ/I = 1.10 N·m / 0.385 m² = 2.857 rad/s².

The linear acceleration (a) of the bucket is related to the angular acceleration by the equation a = Rα, where R is the radius of the pulley. Plugging in the values, we have a = 0.33 m * 2.857 rad/s² = 0.0867 m/s².

To find the angular velocity (ω) at t = 3.00 s, we can use the equation ω = ω₀ + αt, where ω₀ is the initial angular velocity and t is the time.

Since the pulley starts from rest, ω₀ = 0, and plugging in the values, we get ω = 2.857 rad/s² * 3.00 s = 1.15 rad/s.

Similarly, to find the linear velocity (v) of the bucket at t = 3.00 s, we can use the equation v = v₀ + at, where v₀ is the initial velocity.

Since the bucket starts from rest, v₀ = 0, and plugging in the values, we have v = 0.0867 m/s² * 3.00 s = 0.260 m/s.

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Answer:

A) Sue is lifting the bucket by a force of 49.035 newtons.

B) The bucket has an acceleration of 8.207 meters per square second on the Moon.

Explanation:

A) According to the First Newton's Law, a system is at equilibrium when it is either at rest or travelling at constant velocity. In this case, Sue must exert an external force on the bucket, whose magnitude is equal to the weight of the bucket but direction is opposed to it. By Second Newton's Law, we find that:

\(\Sigma F = F - m\cdot g = 0\) (1)

Where:

\(F\) - Lifting force, measured in newtons.

\(m\) - Mass of the bucket, measured in kilograms.

\(g\) - Gravitational acceleration, measured in meters per square second.

If we know that \(m = 5\,kg\) and \(g = 9.807\,\frac{m}{s^{2}}\), then the lifting force is:

\(F = m\cdot g\)

\(F = (5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\)

\(F = 49.035\,N\)

Sue is lifting the bucket by a force of 49.035 newtons.

B) By the Second Newton's Law, we have the following model:

\(\Sigma F = F-m\cdot g = m\cdot a\) (2)

Where \(a\) is the net acceleration of the bucket, measured in meters per square second.

If we know that \(F = 49.035\,N\), \(m = 5\,kg\) and \(g = 1.6\,\frac{m}{s^{2}}\), then the net acceleration of the bucket is:

\(a = \frac{F}{m} -g\)

\(a = \frac{49.035\,N}{5\,kg}-1.6\,\frac{m}{s^{2}}\)

\(a = 8.207\,\frac{m}{s^{2}}\)

The bucket has an acceleration of 8.207 meters per square second on the Moon.

(a) The force applied by Sue in lifting the bucket at a constant velocity is 49 N.

(b) The acceleration of the bucket when lifted on the moon is 8.2 m/s².

The given parameters;

The force applied by Sue in lifting the bucket at a constant velocity is calculated as;

\(F = m(a + g)\)

\(F= mg\\\\F = 5 \times 9.8\\\\F = 49 \ N\)

The acceleration of the bucket when lifted on the moon with the calculated force is;

\(F = m(a + g)\\\\a+g = \frac{F}{m} \\\\a = \frac{F}{m} - g\\\\a = \frac{49}{5} - 1.6\\\\a = 8.2 \ m/s^2\)

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The amount of heat energy required to raise the temperature of a 6.0-cm-diameter copper sphere from -50°C to 150°C is 7.8 kJ.

Given:Radius (r) = 3 cm, Diameter = 6 cm, Temperature T1 = -50°C, Temperature T2 = 150°C

Now we have to find the amount of heat energy required to raise the temperature from T1 to T2.To find out the amount of heat energy, we will use the following formula:

Q = m × C × ΔT

where

Q = heat energy in Joules, m = mass of the copper sphere in kg, C = specific heat capacity of copper, ΔT = change in temperature

ΔT = T2 - T1= 150°C - (-50°C) = 200°C

The specific heat capacity of copper is 0.385 J/g°C or 385 J/kg°C.

We need to convert the radius to meters and mass to kg. Density of copper is 8.96 g/cm³.

Let's apply this conversion:

Volume of the sphere= 4/3πr³= 4/3 × 3.14 × (0.03 m)³= 1.13 × 10⁻⁴ m³

Density of copper = mass / volume mass of copper = volume × density= 1.13 × 10⁻⁴ m³ × 8.96 × 10³ kg/m³= 1.01 × 10⁻¹ kg

Therefore,

Q = m × C × ΔT= 1.01 × 10⁻¹ kg × 385 J/kg°C × 200°C= 7.8 × 10³ J or 7.8 kJ

Hence, the amount of heat energy required to raise the temperature of a 6.0-cm-diameter copper sphere from -50°C to 150°C is 7.8 kJ.

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Answer:

-to defend territory

-to support allies

-to protect economic interests

Explanation:

100% on edge

A nation most likely uses military force-

To defend the nation and its population from immediate danger, military superiority contributes to the maintenance of peace and stability in areas crucial to national interests and finances international security commitments.

The current method used by countries to obtain emergency recruits during times of conflict is drafting. This act, known as conscription in medieval and ancient periods, frequently involved taking young boys against their will from their homes.

Most of the time, leadership is crucial. The Army upholds national laws, accomplishes national goals, and prevents any nation's aggressive actions in order to maintain peace.

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Answer:

\(given \\ accelaration (a)= 2.5 \frac{m}{ {s}^{2} } \\ initial \: velocity(u) = 21 \frac{m}{s} \\ final \: velocity(v) = 31 \frac{m}{s} \\ we \: want \: find \: time \\ newtons \: equation \: of \: motion \: v = u + at \\ 31 = 21 + 2.5 \times t \\ 31 - 21 = 2.5t \\ 10 = 2.5t \\ \frac{10}{2.5} = t \\ t = 4sec\)

Answer:

work equal to Force times distance

By proving both parts of the theorem bidirectionally, we have established that any subspace of a discrete space is discrete, and any subspace of a trivial space is trivial.

Statement 1: Any subspace of a discrete space is discrete.

Proof:

Let's consider a discrete space, which is a space where every subset is an open set. Now, suppose we have a subspace of this discrete space, which means we are considering a subset of the discrete space with the subspace topology induced by the discrete space.

To show that the subspace is also discrete, we need to prove that every subset of the subspace is an open set in the subspace topology.

Let S be a subset of the subspace. Since the subspace inherits its topology from the original discrete space, every subset of the subspace is an intersection of an open set in the discrete space with the subspace.

Since the original discrete space has the property that every subset is open, any intersection of an open set with the subspace will also be open in the subspace topology. Therefore, S is an open set in the subspace topology.

Since this holds for an arbitrary subset S of the subspace, we can conclude that any subspace of a discrete space is discrete.

Now, let's move on to the second statement:

Statement 2: Any subspace of a trivial space is trivial.

Proof:

A trivial space is a space with only one point, where the only open set is the entire space itself.

Suppose we have a subspace of this trivial space. Since the subspace is a subset of the trivial space, it can have at most one point.

To prove that the subspace is trivial, we need to show that the only open set in the subspace is the entire subspace itself.

Since the subspace inherits its topology from the original trivial space, the only open set in the subspace topology will be the intersection of the trivial space with the subspace.

However, since the subspace can have at most one point, the intersection will either be the entire subspace (if it contains the point) or the empty set (if it doesn't contain the point).

In either case, the only open set in the subspace topology is either the entire subspace or the empty set, confirming that the subspace is trivial.

Therefore, we have proved that any subspace of a trivial space is trivial.

By proving both parts of the theorem bidirectionally, we have established that any subspace of a discrete space is discrete, and any subspace of a trivial space is trivial.

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The dishwasher would need to transfer 480 kJ of energy for 400 seconds, or 6 minutes and 40 seconds.

Power is determined by the product of force, velocity, and distance traveled (in the force's direction). For those looking to enter the teaching profession in the state of Uttar Pradesh, this is a fantastic opportunity. Voltage and current together produce electrical power P=VXI.

time = energy / power

480 kJ = 480,000 J

time = energy / power

time = 480,000 J / 1200 W

time = 400 seconds

Hence, for the dishwasher to transfer 480 kJ of energy, it would take 400 seconds, or 6 minutes, 40 seconds.

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Answer:

Then solve for v as a function of t. This is the first equation of motion . It's written like a polynomial — a constant term (v0) followed by a first order term (at). Since the highest order is 1, it's more correct to call it a linear function.

Explanation:

The force of gravitational attraction between the man and woman is 1.0672 × 10⁻⁶ N, if the mass of the man & woman is 80 kg & 50 kg respectively.

Let the Mass of the man, m₁ = 80 kg

Mass of the woman, m₂ = 50 kg

Distance between them, d = 0.5 m

We know the universal gravitational constant, G = 6.67 × 10⁻¹¹ Nm²/kg²

According to Newton's law of gravitation, Gravitational force between two objects having mass m₁ & m₂ at a distance apart d meter, is mathematically formulated as, F = Gm₁m₂/d²

F = (6.67 × 10⁻¹¹ × 80 × 50)/0.5²

F = (2.668 × 10⁻⁷) /0.25

F = 1.0672 × 10⁻⁶ N

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The most likely cause of the electrical failure in satellites and the disruption of radio communications at high latitudes would be a geomagnetic storm.

A geomagnetic storm is a disturbance in the Earth's magnetic field caused by solar wind interactions with the magnetosphere.

During a geomagnetic storm, charged particles from the Sun, known as solar energetic particles, can cause electrical disruptions in satellites by inducing currents in their systems.

These particles can also affect radio communications by interfering with the ionosphere, particularly at high latitudes where the magnetic field lines converge.

Solar flares, which are intense bursts of radiation from the Sun, and coronal mass ejections (CMEs), which are massive eruptions of plasma and magnetic field, can trigger geomagnetic storms.

However, the given information does not specify any indications of solar flares or CMEs, making a geomagnetic storm the most probable cause in this scenario.

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Answer:

a- By kicking the ball into the goal net

Explanation:

You cant pick the ball up so b is not the answer and there is no hoop so c is also not the answer. The answer is A

Answer:

\(\mathbf{\dfrac{I(5l)}{I(l) } =10^{-8}}\)

Explanation:

We are being told that the current is proportional to the tunneling coefficient\(I(l) = I_0 e^{-2kl}\) ;

where l = distance between the tip and the surface.

Let      \(I(l) = I_0 e^{-2kl}\)        ------------  equation (1)

and     \(I(5l) = I_0 e^{-2k(5l)}\)   ------------ equation (2)

Dividing equation (2) by (1); we have :

\(\dfrac{I(5l)}{I(l) } = \dfrac{I_0 e^{-2k(5l)}}{ I_0 e^{-2kl}}\)

\(\dfrac{I(5l)}{I(l) } =e^{-2k(5-1)l}\)

\(\dfrac{I(5l)}{I(l) } =(e^{-2kl})^4\)

where ;

\((e^{-2kl})\) represents the transmission coefficient T = 0.01

Thus; replacing the value for 0.01;we have;

\(\dfrac{I(5l)}{I(l) } =0.01^4\)

\(\mathbf{\dfrac{I(5l)}{I(l) } =10^{-8}}\)

Answer:

Jaliba Kuyateh is the most famous and wealthiest Gambian musician.

Given:

Pressure, P = 300 kPa

Volume, V = 100 cm³

Temperature, T = 30°C

Let's fin the pressure of the gas inside the container if it is heated to 100°C.

Apply the Gay-Lussac's law:

Where:

P1 = 300 kPa

T1 = 30 + 273 = 303 K

T2 = 100 + 273 = 373 K

V1 = V2 (since the container is sealed).

Let's solve for P2.

Rewrite the formula for P2:

Therefore, the pressure if the container is heated to 100°C is 369.31 kPa.

ANSWER:

369.31 kPa

Answer:

False.

Explanation:

Mass is the amount of matter in a given object. Weight is the force acting on an object due to gravity.

Answer:

False

Explanation:

Weight is the act of pulling an object towards the earth surface so weight is a force to amount

Answer:

c

Explanation:

The maximum speed of the object is 1.28 m/s.

What is Speed?

Speed is described as the rate at which an object's position changes in any direction. Speed is defined as the distance traveled divided by the travel time. Speed is a scalar quantity because it only has a direction and no magnitude.

One meter per second is the speed of an object, for instance, if it starts at the origin and moves three meters in three seconds. Simple math can be used to calculate speed. time-distance ratio.

According to the given information:

Force constant K:78.5 n/m

Mass: 0.185 kg

The maximum speed of the object is,

\($$\begin{aligned}v_{\mathbf{n} E} & =A \omega \\& =A \sqrt{\frac{k}{m}}\end{aligned}$$\)

Here A is the amplitude, k is force constant, m is mass.

\($$\begin{aligned}v_{\mathrm{ne:}} & =\left(6.25 \times 10^{-2} \mathrm{~m}\right) \sqrt{\frac{78.5 \mathrm{~N} / \mathrm{m}}{0.185 \mathrm{~kg}}} \\& =1.28 \mathrm{~m} / \mathrm{s}\end{aligned}$$\)

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Answer:

true blood season and answers the question of whether you are not the intended recipient you are not the intended recipient you are not the intended recipient

Answer:

21

Explanation: its actually 20.85 but i guess they round to 21

Explanation:

Mass =150g = 0.15kg

Volume= 35cm^3 = 0.35m^3

Density= Mass×Volume

= 0.15kg × 0.35m^3

= 0.05 kg m^3

The final velocity of the boy/boat system is 1.43 m/s forward. The correct answer is 1.43 m/s forward.

The final velocity of the boy/boat system can be determined using the principle of conservation of momentum. Momentum is defined as the product of an object's mass and velocity. According to the conservation of momentum, the total momentum before the jump is equal to the total momentum after the jump.

Given:
Mass of the boy (m1) = 50 kg
Initial velocity of the boy (v1) = 5 m/s
Mass of the boat (m2) = 125 kg
Initial velocity of the boat (v2) = 0 m/s

Step 1: Calculate the initial momentum of the boy and the boat separately.
Initial momentum of the boy (p1) = m1 * v1
Initial momentum of the boat (p2) = m2 * v2
Since the boat is stationary, its initial momentum is zero.

Step 2: Calculate the total initial momentum of the system.
Total initial momentum (p_initial) = p1 + p2

Step 3: Calculate the final momentum of the system.
Final momentum of the system (p_final) = total initial momentum (p_initial)

Step 4: Use the final momentum to calculate the final velocity of the system.
Final velocity of the system (v_final) = p_final / (m1 + m2)

Substitute the given values into the equation:
v_final = p_final / (m1 + m2)

v_final = (p1 + p2) / (m1 + m2)

v_final = (m1 * v1 + m2 * v2) / (m1 + m2)

Substitute the values:
v_final = (50 kg * 5 m/s + 125 kg * 0 m/s) / (50 kg + 125 kg)

v_final = (250 kg·m/s + 0 kg·m/s) / (175 kg)

v_final = 250 kg·m/s / 175 kg

v_final = 1.43 m/s

So, the final velocity of the boy/boat system is 1.43 m/s forward.

Therefore, the correct answer is "1.43 m/s, forward."

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Answer:

the answer is 5 electrons

Explanation:

because its the same name as the amount of protons

This means that the optimal order quantity for the jewelry firm is 775 stones per order. This would result in a total cost of $40,781.25 for the year (775 stones x $8 per stone + $48 ordering cost x 10 orders per year).

To determine the optimal order quantity for the jewelry firm, we need to consider the trade-off between ordering costs and inventory holding costs. Based on the given information, the annual usage rate of stones is 25 x 200 = 5000 stones.

If the jewelry firm orders 600 stones or more, they can get a price of $8 per stone, which would result in a cost of $8 x 5000 = $40,000 for the year. However, if they order fewer stones, the cost per stone increases. If they order between 400-599 stones, the cost per stone is $9, resulting in a cost of $45,000 for the year. If they order fewer than 400 stones, the cost per stone is $10, resulting in a cost of $50,000 for the year.

To determine the optimal order quantity, we can use the economic order quantity (EOQ) formula, which is:

EOQ = sqrt((2DS)/H)

where D is the annual demand (5000 stones), S is the ordering cost ($48), and H is the holding cost per unit (the difference between the cost of ordering 600 stones and 400 stones, which is $1 per stone).

Plugging in the numbers, we get:

EOQ = sqrt((2 x 5000 x 48)/1)
EOQ = 774.6

This means that the optimal order quantity for the jewelry firm is 775 stones per order. This would result in a total cost of $40,781.25 for the year (775 stones x $8 per stone + $48 ordering cost x 10 orders per year).

Overall, by using the EOQ formula, the jewelry firm can optimize their ordering strategy and minimize their costs.

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I think its "a dropped penny sinks at the bottom of a pond". Because, non-contact force is a force that you don't touch, like gravity or weight, that falls but you didn't drop it on purpose nature did or gravity itself did.